I'm having a hard time answering this question :
Find all the differentiable functions $f$ such as: $\forall (x;y)\in R : f(x^4+y) = x^3f(x) + f(y)$
I would love to get a hint about how to start since this is my first time answering this type of questions.
Thanks a lot.
Differentiate $f(x^4+y)$ by $x$, to get
$$\partial_x f(x^4+y)=4x^3f'(x^4+y)\overset!=\partial_x(x^3f(x)+f(y))=3x^2f(x)+x^3f'(x)$$
Now note that this does not depend on $y$. If $x\neq0$ and you chose $y=z-x^4$ for some $z$ you find $$f'(z)=\frac{3x^2f(x)+x^3f'(x)}{4x^3}$$ so $f'(z)$ is some expression that doesn't even depend on $z$. This means that it has to be constant, and it follows $f(x)=ax+b$ for some constants $a,b$. If you plug it into the initial condition you will find out what these constants are.
I haven't done a detailed analysis and proof, but on first look I am finding that a pure linear (not affine) function $f(u) = au$ where $a \in R$ is about the only one that will fit such a strict condition.
If $f(u) = au$, then $f(x^4 + y) = a(x^4 + y) = a x^4 + ay = x^3 * ax + ay = x^3 f(x) + f(y)$
If you can find anything else that works let me know.